The sum and sum of square corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below: sum_(i=1)^(50)x_(i)=212, sum_(i=1)^(50)x_(i)^(2)=902.8, sum_(i=1)^(50)y_(i)=261 sum_(i=1)^(50)y_(i)^(2)=1457.6 Which is more varying , the length or weight?

The sum and sum of square corresponding to length x (in cm) and weight

1.	The sum and sum of square corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below: sum_(i=1)^(50)x_(i)=212, sum_(i=1)^(50)x_(i)^(2)=902.8, sum_(i=1)^(50)y_(i)=261 sum_(i=1)^(50)y_(i)^(2)=1457.6 Which is more varying , the length or weight?
Answer» SOLUTION :For length, `sumx_(i)=212, sumx_(i)^(2)=902.8,N=50` `:.` MEAN `barx=(sumx_(i))/N=212/50=4.24cm` Variance `=(sumx_(i)^(2))/N-((sumx_(i))/N)^(2)=902.8/50-(212/50)^(2)` `=(45140-44944)/(50^(2))=196/2500` `implies` Standard deviation `SIGMA=14/50=0.28cm` Coefficient of variation of length `=(sigma)/x XX 100` `=0.28/4.24xx 100=6.6` For weight, `sumy_(i)=261,sumy_(i)^(2)=1457.6, N=50` `:.` Mean `=(sumy_(i))/N=261/N=5.22` Variance `sigma^(2)=(sumy_(i)^(2))/N-((sumy_(i))/N)^(2)=1457.6/50-(261/60)^(2)` `=(72880-68121)/((50)^(2))=4759/25500` `implies` Standard deviation `sigma=sqrt(4759/2500)=68.98/50=1.38` Now coefficient of variation of weight `=(sigma)/x xx100=1.38/5.22xx 100=26.43` Therefore, the weight of products are more varying.